# set5 - \$ Set 5 First Order ODEs Part 4 Kyle A Gallivan...

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a39 a38 a36 a37 Set 5: First Order ODEs - Part 4 Kyle A. Gallivan Department of Mathematics Florida State University Ordinary Differential Equations Fall 2009 1

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a39 a38 a36 a37 Linear First Order ODEs Theorem 5.1 (Textbook page 68) . If p ( t ) and g ( t ) are continuous on an open interval I : α < t < β then there exists a unique solution y = φ ( t ) on I to the initial value problem y + p ( t ) y = g ( t ) y ( t 0 ) = y 0 , α < t 0 < β for any value of y 0 . 2
a39 a38 a36 a37 Linear First Order ODEs Corollary 5.2. The unique solution y = φ ( t ) on I to the initial value problem under the conditions of Theorem 5.1 is given by y ( t ) = 1 μ ( t ) bracketleftbig y 0 + integraldisplay t t 0 μ ( s ) g ( s ) ds bracketrightbig μ ( t ) = e z ( t ) z ( t ) = integraldisplay t t 0 p ( t ) dt (Note that μ ( t 0 ) = 1 .) 3

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a39 a38 a36 a37 Linear First Order ODEs Corollary 5.3. All possible solutions to the differential equation on I under the conditions of Theorem 5.1 are given by the general solution y ( t ) = 1 μ ( t ) bracketleftbig C + integraldisplay t t 0 μ ( s ) g ( s ) ds bracketrightbig μ ( t ) = e z ( t ) z ( t ) = integraldisplay t t 0 p ( t ) dt 4
a39 a38 a36 a37 Linear First Order ODEs Assuming the conditions of Theorem 5.1: A general solution in explicit form is known that characterizes all solutions to the ODE. A particular solution to an IVP results by setting C = y 0 The solution requires only two antiderivatives. Possible points of discontinuity or singularity of the solution y ( t ) can be identifed from p ( t ) and g ( t ) . The conditions given are sufficient not necessary. It is possible for the solution y ( t ) to be continuous even when p ( t ) and/or g ( t ) are/is discontinuous.

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